Applied Calculus (with Infotrac) 3rd Edition By Waner, Stefan; Costenoble, Steven Published By Brooks Cole Hardcover
Applied Calculus (with Infotrac) 3rd Edition By Waner, Stefan; Costenoble, Steven Published By Brooks Cole Hardcover
3rd Edition
ISBN: 9781319048518
Author: Colin Adams Jon Rogawski
Publisher: Macmillan Higher Education
Question
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Chapter 15.1, Problem 53E
To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, f(x)=exeaxx, though not defined at x = 0, can be made continuous by assigning the value f(0)=a1

Expert Solution
Check Mark

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, f(x)=exeaxx, though not defined at x = 0, can be made continuous by assigning the value f(0)=a1is derived.

Explanation of Solution

Explanation:

Given: f(x)=exeaxx for a>0

Calculation:

Step 1: The function f(x)=exeaxx,f(0)=a1 is continuous if limx0f(x)=f(0)=a1.

Step 2: We verify this limit using L'Hopital Rule:

limx0exeaxx=limx0ex+aeax1=1+a=a1

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, f(x)=exeaxx, though not defined at x = 0, can be made continuous by assigning the value f(0)=a1 is mathematically proved.

To determine

(b)

The mathematical proof of the statement stating that |f(x)|ex+eax for x>1 using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Expert Solution
Check Mark

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that |f(x)|ex+eax for x>1 using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: x>1, I(a)=0exeaxxdx, f(x)=exeaxx, a > 0

Calculation:

Step 1: We now show that the following integral converges:

I(a)=0exeaxxdx (a>0)

Since, exeax<ex+eax then exeaxx<ex+eaxx for x > 0

If x > 1 we have,

exeaxx<ex+eaxx<ex+eax

That is for x > 1

f(x)<ex+eax

Step 2: Also, since eaxex<eax+ex we have for x > 1

eaxexx<eaxexx<eax+ex

Thus, we get

f(x)<ex+eax

Step 3: Hence, from Step 1 and Step 2, we get

0|f(x)|ex+eax

Step 4: We now show that the integral of the right hand side converges:

0(ex+eax)dx=limR0R(ex+eax)dx=limR(exeaxa|x=0R)=limR(eReaRa+e0+e0a)=limR(eReaRa+1+1a)=1+1a

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

0(ex+eax)xdx also converges for a > 0.

Conclusion: Both the statement stating that |f(x)|ex+eax for x>1 using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

To determine

(c)

The mathematical proof of the equation I(a)=01aexydydx

Expert Solution
Check Mark

Answer to Problem 53E

Solution: The mathematical proof of the equation I(a)=01aexydydx is derived

Explanation of Solution

Given: I(a)=01aexydydx, a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

1aexydy=1xexy|y=1a=1x(exaex.1)=exeaxx

Step 2: Hence,

I(a)=01aexydydx=0(1aexydy)dx=0(exeaxx)dx=I(a)

Conclusion: The equation I(a)=01aexydydx is mathematically proved.

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

I(a)=lnalimT1aeTyydy

Expert Solution
Check Mark

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

I(a)=lnalimT1aeTyydy is derived.

Explanation of Solution

Given: I(a)=0T1aexydydx, a>0

Calculation:

Step 1: By the definition of the improper integral,

I(a)=limT0T1aexydydx

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

0T1aexydydx=1a0Texydxdy=1a(0Texydx)dy=1a(1yexy|x=0T)dy=1a(1y(eTye0.y))dy=1a1eTyydy=1adyy1aeTyydy=lny|1a1aeTyydy=lnaln11aeTyydy=lna1aeTyydy

Combining with Step 1, we get,

I(a)=lnalimT1aeTyydy

Conclusion: By interchanging the order of integration, the equation I(a)=lnalimT1aeTyydy is mathematically proved.

To determine

(e)

The mathematical proof of the statement stating the limit in I(a)=lnalimT1aeTyydy is zero by using the Comparison Theorem.

Expert Solution
Check Mark

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in I(a)=lnalimT1aeTyydy is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: I(a)=lnalimT1aeTyydy, a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: a1 then in the interval of integration y1. As T, we may assume that T > 0

Thus,

eTyyeT.11=eT

Hence,

01aeTyydy1aeTdy=eT(a1)

By the limit limTeT(a1)=0 and the Squeeze Theorem, we conclude that,

limT1aeTyydy=0

Case 2: 0<a<1.

Then,

1aeTyydy=a1eTyydy

and in the interval of integration ay1, therefore

eTyyeTaa(the function eTyyis decreasing).

Hence,

0a1eTyydya1eTaady=1aaeTa

By the limit limT1aaeTa=0 and the Squeeze Theorem, we conclude that

limT1aeTyy=limTa1eTyy=0

We thus showed that for all a > 0, limT1aeTyy=0

Conclusion: The statement stating the limit in I(a)=lnalimT1aeTyydy is zero by using the Comparison Theorem is mathematically proved.

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Chapter 15 Solutions

Applied Calculus (with Infotrac) 3rd Edition By Waner, Stefan; Costenoble, Steven Published By Brooks Cole Hardcover

Ch. 15.1 - Prob. 5ECh. 15.1 - Prob. 6ECh. 15.1 - Prob. 7ECh. 15.1 - Prob. 8ECh. 15.1 - Prob. 9ECh. 15.1 - Prob. 10ECh. 15.1 - Prob. 11ECh. 15.1 - Prob. 12ECh. 15.1 - Prob. 13ECh. 15.1 - Prob. 14ECh. 15.1 - Prob. 15ECh. 15.1 - Prob. 16ECh. 15.1 - Prob. 17ECh. 15.1 - Prob. 18ECh. 15.1 - Prob. 19ECh. 15.1 - Prob. 20ECh. 15.1 - Prob. 21ECh. 15.1 - Prob. 22ECh. 15.1 - Prob. 23ECh. 15.1 - Prob. 24ECh. 15.1 - Prob. 25ECh. 15.1 - Prob. 26ECh. 15.1 - Prob. 27ECh. 15.1 - Prob. 28ECh. 15.1 - Prob. 29ECh. 15.1 - Prob. 30ECh. 15.1 - Prob. 31ECh. 15.1 - Prob. 32ECh. 15.1 - Prob. 33ECh. 15.1 - Prob. 34ECh. 15.1 - Prob. 35ECh. 15.1 - Prob. 36ECh. 15.1 - Prob. 37ECh. 15.1 - Prob. 38ECh. 15.1 - Prob. 39ECh. 15.1 - Prob. 40ECh. 15.1 - Prob. 41ECh. 15.1 - Prob. 42ECh. 15.1 - Prob. 43ECh. 15.1 - Prob. 44ECh. 15.1 - Prob. 45ECh. 15.1 - Prob. 46ECh. 15.1 - Prob. 47ECh. 15.1 - Prob. 48ECh. 15.1 - Prob. 49ECh. 15.1 - Prob. 50ECh. 15.1 - Prob. 51ECh. 15.1 - Prob. 52ECh. 15.1 - Prob. 53ECh. 15.2 - Prob. 1PQCh. 15.2 - 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Prob. 2CRECh. 15 - Prob. 3CRECh. 15 - Prob. 4CRECh. 15 - Prob. 5CRECh. 15 - Prob. 6CRECh. 15 - Prob. 7CRECh. 15 - Prob. 8CRECh. 15 - Prob. 9CRECh. 15 - Prob. 10CRECh. 15 - Prob. 11CRECh. 15 - Prob. 12CRECh. 15 - Prob. 13CRECh. 15 - Prob. 14CRECh. 15 - Prob. 15CRECh. 15 - Prob. 16CRECh. 15 - Prob. 17CRECh. 15 - Prob. 18CRECh. 15 - Prob. 19CRECh. 15 - Prob. 20CRECh. 15 - Prob. 21CRECh. 15 - Prob. 22CRECh. 15 - Prob. 23CRECh. 15 - Prob. 24CRECh. 15 - Prob. 25CRECh. 15 - Prob. 26CRECh. 15 - Prob. 27CRECh. 15 - Prob. 28CRECh. 15 - Prob. 29CRECh. 15 - Prob. 30CRECh. 15 - Prob. 31CRECh. 15 - Prob. 32CRECh. 15 - Prob. 33CRECh. 15 - Prob. 34CRECh. 15 - Prob. 35CRECh. 15 - Prob. 36CRECh. 15 - Prob. 37CRECh. 15 - Prob. 38CRECh. 15 - Prob. 39CRECh. 15 - Prob. 40CRECh. 15 - Prob. 41CRECh. 15 - Prob. 42CRECh. 15 - Prob. 43CRECh. 15 - Prob. 44CRECh. 15 - Prob. 45CRECh. 15 - Prob. 46CRECh. 15 - Prob. 47CRECh. 15 - Prob. 48CRECh. 15 - Prob. 49CRECh. 15 - Prob. 50CRECh. 15 - Prob. 51CRECh. 15 - Prob. 52CRECh. 15 - Prob. 53CRECh. 15 - Prob. 54CRECh. 15 - Prob. 55CRECh. 15 - Prob. 56CRECh. 15 - Prob. 57CRECh. 15 - Prob. 58CRECh. 15 - Prob. 59CRECh. 15 - Prob. 60CRECh. 15 - Prob. 61CRE
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